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BEAMS: SHEARING STRESS
Slide No. 1
Shearing Stress in Beams
Shear and Bending– Although it has been convenient to restrict
the analysis of beams to pure bending, this type of loading is rarely encountered in an actual engineering problem.
– It is much common for the resultant internal forces to consist of a bending moment and shear force.
2
Slide No. 2
Shearing Stress in Beams
Shear and Bending
Pure Bending Bending and Shear Force
Slide No. 3ENES 220 ©AssakkafShear and Bending
( )
( ) 00
0
00
=∫ −==∫=
=∫=−=∫=
=∫ −==∫=
xzxzz
xyxyy
xyxzxxx
yMdAF
dAzMVdAF
dAzyMdAF
στ
στ
ττσ
• Distribution of normal and shearing stresses satisfies
• Transverse loading applied to a beam results in normal and shearing stresses in transverse sections.
• When shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces
• Longitudinal shearing stresses must exist in any member subjected to transverse loading.
3
Slide No. 4
Shearing Stress in Beams
Shear and Bending– The presence of a shear force indicates a
variable bending moment in the beam.– The relationship between the shear force
and the change in bending moment is given by
dxdMV = (42)
Slide No. 5
Shearing Stress in Beams
Shear and Bending– Strictly speaking, the presence of shear
force and resulting shear stresses and shear deformation would invalidate some of our assumption in regard to geometry of the the deformation and the resulting axial strain distribution.
– Plane sections would no longer remain plane after bending, and the geometry
4
Slide No. 6
Shearing Stress in Beams
Shear and Bendingof the actual deformation would become considerably more involved.
– Fortunately, for a beam whose length is large in comparison with the dimensions of the cross section, the deformation effect of the shear force is relatively small; and it is assumed that the longitudinal axial strains are still distributed as in pure bending.
Slide No. 7
Shearing Stress in Beams
Shear and Bending– When this assumption is made, the load
stress relationships developed previously are considered valid.
– The question now being asked:
When are the shearing effects so large thatthey cannot be ignored as a designconsideration?
5
Slide No. 8
Shearing Stress in Beams
Shear and Bending– It is somehow difficult to answer this
question. – Probably the best way to begin answering
this question is to try to approximate the shear stresses on the cross section of the beam.
Slide No. 9
Shearing Stress in Beams
Shearing Stress due to Bending– Suppose that a beam is constructed by
stacking several slabs or planks on top of another without fastening them together.
– Also suppose this beam is loaded in a direction normal to the surface of these slabs.
6
Slide No. 10
Shearing Stress in Beams
Shearing Stress due to Bending
P
Figure 22
(a) Unloaded Stack of Slabs (b) Unglued Slabs loaded
Slide No. 11
Shearing Stress due to Bending
P
Figure 22 (cont’d)
(c) Glued Slabs Unloaded (d) Glued Slabs loaded
Shearing Stress in Beams
7
Slide No. 12
Shearing Stress in Beams
Shearing Stress due to Bending– When a bending load is applied, the stack
will deform as shown in Fig. 22a.– Since the slabs were free to slide on one
one another, the ends do not remain even but staggered.
– Each of the slabs behaves as independent beam, and the total resistance to bending of n slabs is approximately n times the resistance of one slab alone.
Slide No. 13
Shearing Stress in Beams
Shearing Stress– If the slabs of Fig. 22b is fastened or glued,
then the staggering or relative longitudinal movement of slabs would disappear under the action of the force. However, shear forced will develop between the slabs.
– In this case, the stack of slabs will act as a solid beam.
– The fact that this solid beam does not
8
Slide No. 14
Shearing Stress in Beams
Shearing Stressexhibit this relative movement of longitudinal elements after the slabs are glued indicates the presence of shearing stresses on longitudinal planes.
– Evaluation of these shearing stresses will be determined in the next couple of viewgraphs.
Slide No. 15
Shearing Stress in Beams
Development of Shear Stress FormulaConsider the free-body diagram of the short portion of the beam of Figs. 23 and 24a with a rectangular cross section shown in Fig 24b.From this figure,
dytdA =
dAdF σ=
(43)
(44)
9
Slide No. 16
Shearing Stress in Beams
Development of Shear Stress Formula
P
A BC D
x +∆x
x
∆x
tA B
C D
Figure 23
Slide No. 17
Shearing Stress in Beams
Development of Shear Stress Formula
A B
C D
V + ∆V
M + ∆MM V
∆x
yy1c
dy
t
F1 F2
VL VR
VH = τ t ∆x
Figure 24
(a) (b)
(c)
10
Slide No. 18
Shearing Stress in Beams
Development of Shear Stress FormulaThe resultant of these differential forces is
integrated over the area of the cross section, where σ is the flexural stress at a distance y from the neutral axis (surface) and is given by
∫= dAF σ
IMy
−=σ (45)
Slide No. 19
Shearing Stress in Beams
Development of Shear Stress FormulaTherefore, the resultant normal force F1 on the left end of the segment from y1 to the top of the beam is
Similarly, the resultant F2 on the right side of the element is
( )∫∫ −=−=c
y
dytyI
MdAyI
MF1
1
( )∫∫∆+
−=∆+
−=c
y
dytyI
MMdAyI
MMF1
2
(46)
(47)
11
Slide No. 20
Shearing Stress in Beams
Development of Shear Stress FormulaIn reference to Fig. 24c, a summation of forces in the horizontal direction yields
( ) ( )
( )∫
∫∫
∆−=
−∆+
−=
−=
c
y
c
y
c
y
H
dytyIM
dytyI
MdytyI
MM
FFV
1
11
12
(47)
Slide No. 21
Shearing Stress in Beams
Development of Shear Stress FormulaThe average shearing stress τavg is the horizontal force VH divided by the horizontal shear area As = t ∆x between section A and B. Thus
( ) ( )∫∆∆
−==c
ys
H dytyxtI
MAV
1
avgτ (48)
12
Slide No. 22
Shearing Stress in Beams
Development of Shear Stress FormulaIn the limit as ∆x approaches zero, we have
( ) ( )
( )
dytyItdx
dM
dytyItx
M
dytyxtI
M
c
y
c
yx
c
yx
1
1lim
lim
1
1
1
0
0
∫
∫
∫
−=
−
∆∆
=
∆∆
−=
→∆
→∆τ
(49)
Slide No. 23
Shearing Stress in Beams
Development of Shear Stress Formula– Recall that equation 42 relates the bending
moment with the shear force as V = dM/dx. In other words, the shear force V at the beam section where the stress is to be evaluated is given by Eq. 42.
– The integral of Eq. 49 is called the first moment of the area.
∫c
y
dyty1
13
Slide No. 24
Shearing Stress in Beams
Development of Shear Stress Formula– The integral is usually given the symbol
Q. Therefore, Q is the first moment of the portion of the cross-sectional area between the transverse line where the stress is to be evaluated and the extreme fiber of the beam.
∫c
y
dyty1
∫=c
y
dytyQ1
(50)
Extreme Fiber
Extreme Fiber
N.A
Slide No. 25ENES 220 ©Assakkaf
Shear on the Horizontal Face of a Beam Element• Consider prismatic beam
• For equilibrium of beam element( )
∫−
=∆
∑ ∫ −+∆==
A
CD
ADDx
dAyI
MMH
dAHF σσ0
xVxdx
dMMM
dAyQ
CD
A
∆=∆=−
∫=• Note,
flowshearI
VQxHq
xI
VQH
==∆∆
=
∆=∆
• Substituting,
14
Slide No. 26ENES 220 ©Assakkaf
Shear on the Horizontal Face of a Beam Element
flowshearI
VQxHq ==∆∆
=
• Shear flow,
• where
section cross fullofmoment second
above area ofmoment first
'
21
=
∫=
=
∫=
+AA
A
dAyI
y
dAyQ
• Same result found for lower area
HH
qIQV
xHq
∆−=′∆
==′+
′−=′
=∆
′∆=′
axis neutral torespect h moment witfirst
0
Slide No. 27
Shearing Stress in Beams
Example 12Determine the first moment of area Q for the areas indicated by the shaded areas aand b of Fig. 25.
6 ′′
6 ′′
2 ′′
2 ′′
3 ′′
5.1 ′′
a
b
Figure 25
15
Slide No. 28
Example 12 (cont’d)First, we need to locate the neutral axis from the bottom edge:
( )( ) ( )( ) base from 32472
62626232621 ′′==
×+××++×
=Cy
5 ′′
6 ′′
3 ′′·C
2 ′′
N.A
Shearing Stress in Beams
Slide No. 29
Shearing Stress in Beams
Example 12 (cont’d)The first moments of area Qa and Qb are found as follows:
6 ′′
6 ′′
2 ′′
2 ′′
3 ′′
5.1 ′′
a
b· N.A
5 ′′
( )[ ]
[ ] 3
3
in 25.2065.125.13
in 21235.15
=×
−=
=×−=
b
a
Q
Q
3 ′′
16
Slide No. 30
Shearing Stress in BeamsExample 13
A beam is made of three planks, nailed together. Knowing that the spacing between nails is 25 mm and that the vertical shear in the beam is V = 500 N, determine the shear force in each nail.
SOLUTION:
• Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank.
• Calculate the corresponding shear force in each nail.
Slide No. 31ENES 220 ©AssakkafExample 13 (cont’d)
( )( )
( )( )
( )( )
( )( )46
2
3121
3121
36
m1020.16
]m060.0m100.0m020.0
m020.0m100.0[2
m100.0m020.0
m10120
m060.0m100.0m020.0
−
−
×=
×+
+
=
×=
×==
I
yAQ
SOLUTION:
• Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank.
mN3704
m1016.20)m10120)(N500(
46-
36
=
×
×==
−
IVQq
• Calculate the corresponding shear force in each nail for a nail spacing of 25 mm.
mNqF 3704)(m025.0()m025.0( ==
N6.92=F
17
Slide No. 32
Shearing Stress in Beams
Development of Shear Stress FormulaSubstituting for dM/dx of Eq. 42 andof Eq. 50 into Eq. gives
∫=c
y
dytyQ1
( )
ItVQ
QIt
V
dytyItdx
dM c
y
−=
−=
−= ∫
τ
τ
1
1
1
(51)
Slide No. 33
Shearing Stress in Beams
Development of Shear Stress FormulaEq. 51 provides a formula to compute the horizontal (longitudinal) and vertical (transverse) shearing stresses at each point at a beam. These vertical and horizontal stresses are equivalent in magnitude.
x
y
xyτxyτ
yxτ
yxτ
18
Slide No. 34ENES 220 ©Assakkaf
Determination of the Shearing Stress in a Beam• The average shearing stress on the horizontal
face of the element is obtained by dividing the shearing force on the element by the area of the face.
ItVQ
xtx
IVQ
Axq
AH
ave
=
∆∆
=∆∆
=∆∆
=τ
• On the upper and lower surfaces of the beam, τyx= 0. It follows that τxy= 0 on the upper and lower edges of the transverse sections.
• If the width of the beam is comparable or large relative to its depth, the shearing stresses at D1and D2 are significantly higher than at D.
Slide No. 35
Shearing Stress in Beams
19
Slide No. 36
Shearing Stress in Beams
Shearing Stress FormulaAt each point in the beam, the horizontal and vertical shearing stresses are given by
ItVQ
=τWhereV = shear force at a particular section of the beamQ = first moment of area of the portion of the cross-sectional area
between the transverse line where the stress is to be computed.I = moment of inertia of the cross section about neutral axist = average thickness at a particular location within the cross section
(52)
LECTURE 14. BEAMS: SHEARING STRESS (6.1 – 6.4) Slide No. 37
Shearing Stress in Beams
Shearing Stress FormulaHow accurate is the shearing stress formula?
• The formula is accurate if t is not too great.• For a rectangular section having a depth twice
the width, the maximum stress as computed by more rigorous method is about 3% greater than that given by Eq. 52.
• If the beam is square, the error is about 12%.• If the width is four times the depth, the error is
about 100 %.
20
Slide No. 38
Shearing Stress in Beams
Shearing Stress FormulaHow accurate is the shearing stress formula?
Greatt is small
3% errord = 2 t
12% errord = t
100% error, worst case4d = t
d
tt t t
Slide No. 39
Shearing Stress in Beams
Variation of Vertical Shearing Stress in the Cross Section
N.AV
Max Stress
21
Slide No. 40ENES 220 ©Assakkaf
Shearing Stresses τxy in Common Types of Beams• For a narrow rectangular beam,
AV
cy
AV
IbVQ
xy
23
123
max
2
2
=
−==
τ
τ
• For American Standard (S-beam) and wide-flange (W-beam) beams
web
ave
AV
ItVQ
=
=
maxτ
τ
Slide No. 41
Shearing Stress in Beams
Example 14A machine part has a T-shaped cross section and is acted upon in its plane of symmetry by the single force shown. Determine (a) the maximum compressive stress at section n-n and (b) the maximum shearing stress.
22
Slide No. 42ENES 220 ©Assakkaf
Shearing Stress in Beams
Example 14 (cont’d)
15 in
12 in
1.5 kips n4 in
0.5 in
2 in
0.5 in
n
Slide No. 43
Shearing Stress in Beams
Example 14 (cont’d)
15 in
12 in
1.5 kips n
n
1.5 kips
V
M
V (kip)
1.5(-)
(+)M (kip-in)
22.5
n
nMn-n =18
12 in
23
Slide No. 44
Shearing Stress in Beams
Example 14 (cont’d)First, we need to locate the neutral axis. Let’s make our reference from the bottom edge.
( )( ) ( )( )
x
r
C
IyM
yyy
y
max
maxcomten
Stress Max.
in 833.1in 667.0833.15.2
in 833.15.045.02
5.0425.025.021
=
===−=
=×+×
×++×=
4 in
0.5 in
2 in
0.5 in
Slide No. 45
Shearing Stress in BeamsExample 14 (cont’d)
Next find the moment of inertia about the neutral axis:
4 in
0.5 in
2 in
0.5 in
· N.A.1.833 in
C
( ) ( ) ( ) 4333
in 417.13167.05.3
3667.04
3833.15.0
=
−+=xI
(a) Maximum normal stress is a compressivestress:
( ) (C) ksi 3.23417.1833.118max
max === −
IcM nnσ
24
Slide No. 46
Shearing Stress in BeamsExample 14 (cont’d)
(b) Maximum shearing stress:The maximum value of Q occurs at the neutral axis.Since in this cross section the width t is minimum atthe neutral axis, the maximum shearing stress willoccur there. Choosing the area below a-a at theneutral axis, we have
4 in
0.5 in
2 in
0.5 in
N.A.1.833 in
C· ( )( )
( )( ) ksi 778.1
5.0 417.1840.0 5.1
in 840.0833.15.02833.1
max
3
===
==
ItVQ
Q
τ
Slide No. 47ENES 220 ©AssakkafFurther Discussion of the Distribution of Stresses
in a Narrow Rectangular Beam
−= 2
21
23
cy
AP
xyτI
Pxyx +=σ
• Consider a narrow rectangular cantilever beam subjected to load P at its free end:
• Shearing stresses are independent of the distance from the point of application of the load.
• Normal strains and normal stresses are unaffected by the shearing stresses.
• From Saint-Venant’s principle, effects of the load application mode are negligible except in immediate vicinity of load application points.
• Stress/strain deviations for distributed loads are negligible for typical beam sections of interest.
25
Slide No. 48
Shearing Stress in BeamsExample 15
A timber beam is to support the three concentrated loads shown. Knowing that for the grade of timber used,
psi120psi1800 == allall τσ
determine the minimum required depth d of the beam.
SOLUTION:
• Develop shear and bending moment diagrams. Identify the maximums.
• Determine the beam depth based on allowable normal stress.
• Determine the beam depth based on allowable shear stress.
• Required beam depth is equal to the larger of the two depths found.
Slide No. 49
Shearing Stress in Beams
Example 15 (cont’d)
SOLUTION:
Develop shear and bending moment diagrams. Identify the maximums.
inkip90ftkip5.7kips3
max
max⋅=⋅=
=MV
26
Slide No. 50ENES 220 ©AssakkafExample 15 (cont’d)
( )
( ) 2
261
261
3121
in.5833.0
in.5.3
d
d
dbcIS
dbI
=
=
==
=
• Determine the beam depth based on allowable normal stress.
( )in.26.9
in.5833.0in.lb1090psi 1800 2
3
max
=
⋅×=
=
dd
SM
allσ
• Determine the beam depth based on allowable shear stress.
( )in.71.10
in.3.5lb3000
23psi120
23 max
=
=
=
dd
AV
allτ
• Required beam depth is equal to the larger of the two.in.71.10=d
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